components of velocity


by klm
Tags: components, velocity
klm
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#1
Oct12-07, 06:45 AM
P: 167
A ball is thrown toward a cliff of height h with a speed of 30m/s and an angle of 60 degrees above horizontal. It lands on the edge of the cliff 4.0s later.

How high is the cliff?
do you just use the equation h=.5gt^2 ? and put it in -9.80=g and 4=t ?
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Doc Al
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#2
Oct12-07, 07:31 AM
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You also need to consider the initial speed in the y-direction; here's the complete equation:
[tex]y = y_0 + v_{0y} t - (1/2) g t^2[/tex]
klm
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#3
Oct12-07, 07:36 AM
P: 167
would this be correct: y0=26 , v0=30, g=-9.8 , t=4 ... so y=26+(30x4)- .5(-9.8)(4)^2 = 224.4 m ?

i got vy by doing v0sintheta= 26

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#4
Oct12-07, 07:44 AM
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components of velocity


Quote Quote by klm View Post
would this be correct: y0=26 , v0=30, g=-9.8 , t=4 ... so y=26+(30x4)- .5(-9.8)(4)^2 = 224.4 m ?

i got vy by doing v0sintheta= 26
No. In the formula I gave:
y0 is your initial position, which I presume is on the ground at height = 0
v0 (which I'll change to v0y) is the vertical component of the initial velocity, what you call v0sin(theta)
g = 9.8 m/s^2

To make it less confusing, I'll relabel v0 to be v0y in my equation.
klm
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#5
Oct12-07, 07:48 AM
P: 167
ohh sorry, so y=0 + 26(4) - .5(9.8)(4)^2 = 25.6 m
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#6
Oct12-07, 07:55 AM
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Good!
klm
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#7
Oct12-07, 07:59 AM
P: 167
thanks! can you help me with the next part too..! What was the maximum height of the ball?
i think the equation is just the same as the one you wrote, but just cut t=4 in half to get the peak height so t=2 ..so y= 26(2)-.5(9.8)(2)^2 = 32.4 m ?
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#8
Oct12-07, 08:10 AM
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No, you can't assume that the peak is at half the time. After all, it lands up on a cliff, so it spent more time rising than falling. (If it fell back down to the original height, then you'd be correct.)

Instead, use a velocity equation for the y-direction to figure out the time when it reaches maximum height. Hint: At the maximum height, what's the vertical speed?
klm
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#9
Oct12-07, 08:14 AM
P: 167
um i think the vertical speed should be 0
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#10
Oct12-07, 08:16 AM
P: 167
so would it be okay to use the equation vfy= viy +ayT so 0= 26+9.8t so t= 2.65 and then cut then stick that time in that first equation you gave me?
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#11
Oct12-07, 08:18 AM
P: 167
so y= 26(2.65) -.5(9.8)(2.65)^2 =34 .4m ?
klm
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#12
Oct12-07, 08:31 AM
P: 167
are the equations i used alright?
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#13
Oct12-07, 08:38 AM
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Very good!
klm
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#14
Oct12-07, 08:39 AM
P: 167
thank you Doc Al! do you mind one more question, it will be the last one i promise! =)
klm
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#15
Oct12-07, 08:41 AM
P: 167
What is the ball's impact speed?

i thought what you should do is find the final velocity in the x component and y component. and i thought that vfx= 15 since there is no acc in the x direction and vfy= -26 b/c of neg acc. and then i thought you should take the magnitude, but this does not work out to be right. do you know what i am doing wrong

wait i think i did this wrong, look at my next post please
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#16
Oct12-07, 08:45 AM
P: 167
oh no actually should vfy= 65.2 because i tried the equation vfy=viy+ay x T so 26+(9.8x4)= 65.2
so do i do square root (15^2 + 65.2^2) = 66.9 ?
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#17
Oct12-07, 08:56 AM
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Quote Quote by klm View Post
oh no actually should vfy= 65.2 because i tried the equation vfy=viy+ay x T so 26+(9.8x4)= 65.2
Careful here. ay = -9.8 m/s^2.
klm
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#18
Oct12-07, 08:58 AM
P: 167
ohhh so should it be vfy= -13.2 and then do the sqaure root (13.2^2 +15^2 ) =19.98


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